Sets and first order logic question

Question

this question has me stumped. Any help would be greatly appreciated.

It goes like this:

For the domain of all sets, give a first-order logic formula saying - There exists no set such that all other sets are its element.

So far I have this:

Let $x$ and $y$ be all sets. $\lnot\exists x\,\forall\,y\,(y\in x \impliedby x\ne y)$

Note, the reason I've put $x$ does not equal $y$ is because the question says "all OTHER" sets, so I interpreted that as, it does not include itself, so it's not a member of itself but again I'm not sure.

Answer 1

The statement:

"There does not exists a set such that all other sets are its elements."

can be written in first order language of set theory $\mathcal{L} = \{\in\}$ as

$\neg(\exists x)((\forall y)(y \neq x \Rightarrow y \in x) \wedge (x \notin x))$

Answer 2

HINT:

You merely wrote that $x$ is not one of its own elements (if $y$ is a member of $x$, then $y\neq x$). The other direction is the other implication (if $y\neq x$, then $y$ is a member of $x$).